Question

The water gas shift reaction is used commercially to produce H2(g):

CO(g)+H2O(g)⇌CO2(g)+H2(g).

Use the following data to determine:

ΔfH∘[CO2(g)] = -393.5 kJ/mol
ΔfH∘[H2(g)] = 0 kJ/mol
ΔfH∘[CO(g)] = -110.5 kJ/mol
ΔfH∘[H2O(g)] = -241.8 kJ/mol
ΔS∘[CO2(g)] = -393.5 Jmol−1K−1
ΔS∘[H2(g)] = -393.5 Jmol−1K−1
ΔS∘[CO(g)] = -393.5 Jmol−1K−1
ΔS∘[H2O(g)] = -393.5 Jmol−1K−1

a. ΔrH∘ at 298 K.

b. ΔrS∘ at 298 K.

c. ΔrG∘ at 298 K.

d. K at 650 K .

Answer 1

a. For the reaction standard change in enthalpy is:

_rH^o = H^o (products) - H^o (reactants)

= [H^o(CO2) + H^o(H2)] - [H^o(CO) + H^o(H2O)]

= [-393.5+0]-[-110.5-241.8] = -41.2 kJ/mol

b. For the reaction standard change in entropy is:

_rS^o = S^o (products) - S^o (reactants)

= [S^o(CO2) + S^o(H2)] - [S^o(CO) + S^o(H2O)]

= [-393.5-393.5]-[-393.5-393.5] = 0 J/mol.K

c. _rG^o = _rH^o - T_rS^o

= -41.2 kJ/mol

d.  _rG^o = -RT ln K

-41200 = -(8.314*298) ln K

ln K = 16.63

K = 6*10^-8